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[tex] \frac{dy}{dx}= \frac{1}{2} \frac{d}{dx}e^x + \frac{1}{2} \frac{d}{dx}e^{-x}= \frac{1}{2}(e^x-e^{-x}) \\ [/tex].

Therefore [tex] (\frac{dy}{dx})^2=\frac{1}{4}(e^x-e^{-x})\\[/tex] Therefore [tex] 1+(\frac{dy}{dx})^2=\frac{1}{4}(e^{2x} -2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}\\ [/tex].

Therefore [tex] \int_0^{ln4}\frac{1}{2}(e^{2x}-2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}dx[/tex]. Do I solve this by substitution by letting [tex] u^2 = e^{2x}-2e^xe^{-x}+e{-2x}+4[/tex]? Or what way should I approach it? Using the said substitution I get the following integral: [tex] \int u^2 \frac{1}{e^{2x}-e^{-2x}}du[/tex]